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Constants for water:
[tex]\[
\begin{array}{|l|l|}
\hline
\Delta H_{\text{vap}} & 40.65 \, \text{kJ/mol} \\
\hline
\Delta H_f & -285.83 \, \text{kJ/mol} \\
\hline
\Delta H_{\text{fusion}} & 6.03 \, \text{kJ/mol} \\
\hline
\text{specific heat} & 4.186 \, \text{J/g} ^{\circ} \text{C} \\
\hline
\text{molar mass} & 18.02 \, \text{g/mol} \\
\hline
\end{array}
\][/tex]

How much energy is generated from freezing 2.5 g of water?

A. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 4.186 \, \text{kJ/mol}\)

B. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 40.65 \, \text{kJ/mol}\)

C. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times (-285.83) \, \text{kJ/mol}\)

D. [tex]\(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 6.03 \, \text{kJ/mol}\)[/tex]

Sagot :

GinnyAnswer
To determine how much energy is generated from freezing \(2.5 \, \text{g}\) of water, we will proceed step-by-step.

1. Identify the relevant constants:
For freezing water, the relevant enthalpy change is the heat of fusion, \(\Delta H_{\text{fusion}} = 6.03 \, \text{kJ/mol}\).

2. Determine the mass of water:
Given mass of water \( m = 2.5 \, \text{g} \).

3. Calculate the moles of water:
We need to convert the mass of water to moles. The molar mass of water, \( M_{\text{H}_2\text{O}} = 18.02 \, \text{g/mol} \).

[tex]\[ \text{moles of water} = \frac{\text{mass of water}}{\text{molar mass of water}} = \frac{2.5 \, \text{g}}{18.02 \, \text{g/mol}} = 0.1387 \, \text{mol} \][/tex]

4. Calculate the energy using \(\Delta H_{\text{fusion}}\):
[tex]\[ \text{energy generated} = (\text{moles of water}) \times \Delta H_{\text{fusion}} \][/tex]
[tex]\[ \text{energy generated} = 0.1387 \, \text{mol} \times 6.03 \, \text{kJ/mol} = 0.8366 \, \text{kJ} \][/tex]

Therefore, the amount of energy generated from freezing \(2.5 \, \text{g}\) of water is \(0.8366 \, \text{kJ}\).

The closest matching option in your list is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times 6.03 \, \text{kJ/mol} \)

So, the correct choice is:
D. \(2.5 \, \text{g} \times \frac{1 \, \text{mol}}{18.02 \, \text{g}} \times (6.03) \, \text{kJ/mol}
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